If-z-1-and-z-2-are-complex-numbers-such-that-z-2-1-and-z-1-2z-2-2-z-1-z-2-1-then-z-1-is-equal-to-

Question Number 140442 by EnterUsername last updated on 07/May/21

$$\mathrm{If}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{2}} \mid\neq\mathrm{1}\:\mathrm{and} \\ $$$$\mid\left({z}_{\mathrm{1}} −\mathrm{2}{z}_{\mathrm{2}} \right)/\left(\mathrm{2}−{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\mid=\mathrm{1},\:\mathrm{then}\:\mid{z}_{\mathrm{1}} \mid\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\_\_\_\_\_. \\ $$

Answered by mr W last updated on 08/May/21

∣((r_1 e^(θ_1 i) −2r_2 e^(θ_2 i) )/(2−r_1 e^(θ_1 i) r_2 e^(−θ_2 i) ))∣=1 ∣(((r_1 e^((θ_1 −θ_2 )i) −2r_2 )e^(θ_2 i) )/(2−r_1 r_2 e^((θ_1 −θ_2 )i) ))∣=1 ∣(((r_1 e^(θ_3 i) −2r_2 )e^(θ_2 i) )/(2−r_1 r_2 e^(θ_3 i) ))∣=1 ∣(((√((r_1 cos θ_3 −2r_2 )^2 +(r_1 sin θ_3 )^2 ))e^(θ_4 i) e^(θ_2 i) )/( (√((2−r_1 r_2 cos θ_3 )^2 +(−r_1 r_2 sin θ_3 )^2 ))e^(θ_5 i) ))∣=1 ∣((√(r_1 ^2 +4r_2 ^2 −4r_1 r_2 cos θ_3 ))/( (√(4+r_1 ^2 r_2 ^2 −4r_1 r_2 cos θ_3 )))) e^(θ_6 i) ∣=1 ((√(r_1 ^2 +4r_2 ^2 −4r_1 r_2 cos θ_3 ))/( (√(4+r_1 ^2 r_2 ^2 −4r_1 r_2 cos θ_3 ))))=1 r_1 ^2 +4r_2 ^2 =4+r_1 ^2 r_2 ^2 r_1 ^2 (1−r_2 ^2 )=4(1−r_2 ^2 ) r_1 ^2 =4 since r_2 ≠1 ⇒r_1 =2=∣z_1 ∣

$$\mid\frac{{r}_{\mathrm{1}} {e}^{\theta_{\mathrm{1}} {i}} −\mathrm{2}{r}_{\mathrm{2}} {e}^{\theta_{\mathrm{2}} {i}} }{\mathrm{2}−{r}_{\mathrm{1}} {e}^{\theta_{\mathrm{1}} {i}} {r}_{\mathrm{2}} {e}^{−\theta_{\mathrm{2}} {i}} }\mid=\mathrm{1} \\ $$$$\mid\frac{\left({r}_{\mathrm{1}} {e}^{\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right){i}} −\mathrm{2}{r}_{\mathrm{2}} \right){e}^{\theta_{\mathrm{2}} {i}} }{\mathrm{2}−{r}_{\mathrm{1}} {r}_{\mathrm{2}} {e}^{\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right){i}} }\mid=\mathrm{1} \\ $$$$\mid\frac{\left({r}_{\mathrm{1}} {e}^{\theta_{\mathrm{3}} {i}} −\mathrm{2}{r}_{\mathrm{2}} \right){e}^{\theta_{\mathrm{2}} {i}} }{\mathrm{2}−{r}_{\mathrm{1}} {r}_{\mathrm{2}} {e}^{\theta_{\mathrm{3}} {i}} }\mid=\mathrm{1} \\ $$$$\mid\frac{\sqrt{\left({r}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{3}} −\mathrm{2}{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({r}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{3}} \right)^{\mathrm{2}} }{e}^{\theta_{\mathrm{4}} {i}} {e}^{\theta_{\mathrm{2}} {i}} }{\:\sqrt{\left(\mathrm{2}−{r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{3}} \right)^{\mathrm{2}} +\left(−{r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{sin}\:\theta_{\mathrm{3}} \right)^{\mathrm{2}} }{e}^{\theta_{\mathrm{5}} {i}} }\mid=\mathrm{1} \\ $$$$\mid\frac{\sqrt{{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{4}{r}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{3}} }}{\:\sqrt{\mathrm{4}+{r}_{\mathrm{1}} ^{\mathrm{2}} {r}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{3}} }}\:{e}^{\theta_{\mathrm{6}} {i}} \mid=\mathrm{1} \\ $$$$\frac{\sqrt{{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{4}{r}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{3}} }}{\:\sqrt{\mathrm{4}+{r}_{\mathrm{1}} ^{\mathrm{2}} {r}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{3}} }}=\mathrm{1} \\ $$$${r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{4}{r}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{4}+{r}_{\mathrm{1}} ^{\mathrm{2}} {r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${r}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}−{r}_{\mathrm{2}} ^{\mathrm{2}} \right)=\mathrm{4}\left(\mathrm{1}−{r}_{\mathrm{2}} ^{\mathrm{2}} \right) \\ $$$${r}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{4}\:{since}\:{r}_{\mathrm{2}} \neq\mathrm{1} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\mathrm{2}=\mid{z}_{\mathrm{1}} \mid \\ $$

Commented by EnterUsername last updated on 12/May/21

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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